The increasing order of the boiling points for the following compounds is:
(I) C$$_2$$H$$_5$$OH    (II) C$$_2$$H$$_5$$Cl    (III) C$$_2$$H$$_5$$CH$$_3$$    (IV) C$$_2$$H$$_5$$OCH$$_3$$

First we list the four species with their usual names so that we can compare them conveniently.

$$\begin{aligned} (I)\; &C_2H_5OH &&\text{is ethanol (an alcohol).}\\ (II)\; &C_2H_5Cl &&\text{is ethyl chloride (an alkyl halide).}\\ (III)\; &C_2H_5CH_3 &&\text{is }CH_3CH_2CH_3\text{, propane (a hydrocarbon).}\\ (IV)\; &C_2H_5OCH_3 &&\text{is }CH_3OCH_2CH_3\text{, ethyl methyl ether (an ether).} \end{aligned}$$

The boiling point of a liquid is governed by the strength of the intermolecular forces that must be overcome to convert it to vapour. We therefore analyse the type and relative magnitude of the forces present in each compound.

We have three main kinds of intermolecular attractions in organic molecules:

1. Hydrogen bonding (very strong): occurs when an O-H, N-H or F-H bond donates a hydrogen to the lone pair of another electronegative atom.
2. Dipole-dipole interaction (moderate): present in polar molecules that do not possess an H atom bonded directly to O, N or F.
3. London (dispersion) forces (weak but increase with molecular mass and surface area): present in all molecules, dominant in non-polar ones.

Now we consider each molecule in turn.

Ethanol, $$(I)\; C_2H_5OH$$
Ethanol contains an $$O-H$$ bond. According to the first criterion above, it can form intermolecular hydrogen bonds. These are the strongest of the three interactions listed, so ethanol is expected to have the highest boiling point among the four.

Ethyl methyl ether, $$(IV)\; C_2H_5OCH_3$$
The oxygen atom has lone pairs, making the molecule polar, but there is no $$O-H$$ bond. Hence it cannot donate a hydrogen for hydrogen bonding; it exhibits dipole-dipole interactions only. These are weaker than hydrogen bonds, so its boiling point is lower than that of ethanol.

Ethyl chloride, $$(II)\; C_2H_5Cl$$
Because of the electronegativity difference between $$C$$ and $$Cl$$, the $$C-Cl$$ bond is polar, giving the molecule a dipole moment. Thus it also shows dipole-dipole interactions, in addition to somewhat stronger dispersion forces than (IV) because the chlorine atom is heavier and more polarisable. Consequently its boiling point should be a little higher than that of the ether (IV) but still well below that of ethanol (I).

Propane, $$(III)\; C_2H_5CH_3 = CH_3CH_2CH_3$$
Propane is completely non-polar; the only forces holding its molecules together are London dispersion forces. Its molar mass (44 g mol$$^{-1}$$) is close to those of the other compounds, but the absence of polarity or hydrogen bonding makes its boiling point the lowest of the set.

Collecting these qualitative results, we write

London only $$\lt$$ dipole + London $$\lt$$ H-bond + dipole + London

Assigning each molecule to its category gives the numerical inequality

$$BP_{(III)} \lt BP_{(IV)} \lt BP_{(II)} \lt BP_{(I)}$$

or in words,

$$(III)\; C_2H_5CH_3 \; \lt \; (IV)\; C_2H_5OCH_3 \; \lt \; (II)\; C_2H_5Cl \; \lt \; (I)\; C_2H_5OH.$$

This sequence corresponds exactly to Option C in the question.

Hence, the correct answer is Option C.